MBA.Aspirant wrote:When solving standard deviation problems, is it considered the population or the sample one? i.e you divide by n or n-1?
This is an example where the 2 approaches yield different answers
If the standard deviation of 4, 6, 14, 16 and 20 is K, then the variance of 11, 15, 31, 35 and 43 is
a) 4K^2
b) 4K^2 + 3
c) (2K + 3)^2
d) 2K + 3
another lame question: the median of a set of 16 terms, is the average of 8th term and 9th or 7th and 8th?
While solving the problem you provided, I don't think it is necessary to divide by n-1; for that matter I don't see why you are conflicted between the two as there is no indication of a sample.
To solve the problem, you need to know the properties of standard deviation with respect to changes in the original set.
When each number is multiplied by a same constant, then the new standard deviation is found by multiplying the
old S.D. by that same constant. Whereas,
adding a constant to each number won't change the S.D. So, having these in mind, let's delve into the question.
Note that
the second set constitutes numbers which resulted from multiplying the numbers in the first set by 2 and adding 3. Therefore, since S.D. for the first set is K, S.D. for the second set has to be 2K.[Here,
note that adding 3 to all has no impact whatsoever on the S.D.]
But in the question, we are asked for the variance, which is by definition the square of S.D. Therefore, (2K)^2 =
4K^2
In regards to your second question, you have to take the average of
the 8th. and the 9th. terms. To get this, divide the last term by 2 to get the one of the elements and take the next upper element as the other.
Hope this helps.
